The amount of heat -150C ice convert to 00C water
q1 = mcT
= 100*2.060(0-(-15)
= 100*2.060*15
= 3090J
The amount of heat of fusion of ice
q2 = mH fus
= 100*333.5
= 33350J
The amount of heat water convert at 00C to 1000C
q3 = mcT
= 100*4.184*(100-0)
= 41840J
The amount of heat steam convert 1000C to 1100C
q4 = mcT
= 100*2.040*(110-100)
= 2040J
The amount of heat of vaporization
q5 = mH vap
= 100*2260
= 226000J
The total amount of heat
q = q1 + q2 + q3 + q4 + q5
= 3090 + 33350 + 41840 + 2040 + 226000
= 306320J
= 306.32KJ >>>>answer
Exercise #11 (65 points) Specific Heat 100.0 g of ice at -15.0°C become steam at 110°C...
100.0 g of ice at -15.0°C become steam at 110°C. Using the data reported in the tables, calculate the amount of heat necessary for the process to occur at standard pressure. Draw the behavior of heat versus temperature. For full credit, report on the plot all the necessary information. Specific Heat Capacities (J/(g*K)) H2O (s): 2.060 H2O (l), 25°C: 4.184 H2O (g), 100°C: 2.040 Change in Enthalpy of H2O Fusion: 333.5 J/g Vaporization: 2260 J/g
How much heat is required to convert 15.0 g of ice at -11.0 °C to steam at 100.0 °C? Express your answer in (a)joules, (b)calories, and (c)Btu.
How much heat is released when 105 g of steam at 100.0°C is cooled to ice at -15.0°C? The enthalpy of vaporization of water is 40.67 kJ/mol, the enthalpy of fusion for water is 6.01 kJ/mol, the molar heat capacity of liquid water is 75.4 J/(mol • °C), and the molar heat capacity of ice is 36.4 J/(mol • °C). A)347 kJ B)54.8 kJ C)319 kJ D)273 kJ
How much heat (in kJ) is released when 125.0 g of steam at 100.0°C is cooled to ice at -15.0°C? The enthalpy of vaporization of water is 40.67 kJ/mol, the enthalpy of fusion for water is 6.01 kJ/mol, the molar heat capacity of liquid water is 75.4 J/(mol ∙ °C), and the molar heat capacity of ice is 36.4 J/(mol ∙ °C).
6) There is 15.0 g of ice at 0.0C. How many grams of water at 50.0C must be added to the ice to melt all the ice and keep the temperature of the mixture at 0.OC? H2O(s)= 2.06J/gc Specific heat: H2O(0) = (4.18 J/g C) H20 heat of vaporization=2260 J/g H2O(g)=(2.03 J/g C) H2O heat of fusion=333 J/g b)25 8 c)23.98 d)7. 58 )0.428
Given that the specific heat capacities of ice and steam are 2.06 J/g°C and 2.03 J/g°C, the molar heats of fusion and vaporization for water are 6.02 kJ/mol and 40.6 kJ/mol, respectively, and the specific heat capacity of water is 4.18 J/g°C, calculate the total quantity of heat evolved when 24.1 g of steam at 158°C is condensed, cooled, and frozen to ice at -50.°C.
You add 100.0 g of water at 51.0 °C to 100.0 g of ice at 0.00 °C. Some of the ice melts and cools the water to 0.00 °C. When the ice and water mixture reaches thermal equilibrium at 0 °C, how much ice has melted? (The specific heat capacity of liquid water is 4.184 J/g · K. The enthalpy of fusion of ice at 0 °C is 333 J/g.) Mass of ice = References Use the References to access...
How much heat (in kJ) is required to convert 431 g of liquid H2O at 24.0°C into steam at 157°C? (Assume that the specific heat of liquid water is 4.184 J/g·°C, the specific heat of steam is 2.078 J/g·°C, and that both values are constant over the given temperature ranges. The normal boiling point of H2O is 100.0°C.
How much heat (in kJ) is required to convert 431 g of liquid H2O at 23.6°C into steam at 148°C? (Assume that the specific heat of liquid water is 4.184 J/g·°C, the specific heat of steam is 2.078 J/g·°C, and that both values are constant over the given temperature ranges. The normal boiling point of H2O is 100.0°C. The heat of vaporization (ΔHvap) is 40.65 kJ/mol.)
How much heat is required to convert 13.0 g of ice at -12.0 ∘C to steam at 100.0 ∘C? Express your answer in (a)joules, (b)calories, and (c)Btu